Angular velocity of rod and bug

[p37]

Homework Statement

A thin rod has a length of 0.138 m and rotates in a circle on a frictionless tabletop. The axis is perpendicular to the length of the rod at one of its ends. The rod has an angular velocity of 0.221 rad/s and a moment of inertia of 1.08 x 10-3 kg·m2. A bug standing on the axis decides to crawl out to the other end of the rod. When the bug (whose mass is 5 x 10-3 kg) gets where it's going, what is the change in the angular velocity of the rod?

Relevant Equations

Lfinal = Linitial
L=Iw

I(i)w(i)= I(f)w(f)

I(i)= 1.08 x 10-3 kg·m2
w(i)= 0.221 rad/s
I(f)= mr^2 + I(i) = (5 x 10^-3)(.138)^2 + (1.08 x 10^-3)

(1.08 x 10-3)(.221) = ((1.08 x 10^-3)+9.22 x 10^-5))w(f)

w(f) = (2.3868 x 10^-4)/(0.00117522)

w(f)= 0.203094 rad/s

This is my attempt; however, I cannot seem to get it right. Any help on what I am doing wrong?

 

[haruspex]

What answer did you put in?
Note it says "what is the change in"

 

[p37]

I put in 0.203094 but it's wrong
I thought maybe it was .221 - 0.203094 which is .017906 which isn't right either
could it be the negative of that?

I only have one attempt left

 

[haruspex]

I don't like the fact that it specifies the "moment of inertia " of the rod without stating the axis for that. You are left to guess whether it means about the rod's mass centre or about the given axis of rotation.
You are quoting too many decimal places in your answers, so that's another possibility, but most systems don’t police that.
I haven't checked your arithmetic. I will now.

Update: I agree with your numbers.

 

[p37]

Yeah, I am not sure. I followed videos online with similar problems and they did it the same way. Usually, I don't have an issue with sig figs or decimal points on my program.

 

[haruspex]

Yeah, I am not sure. I followed videos online with similar problems and they did it the same way. Usually, I don't have an issue with sig figs or decimal points on my program.

I can only suggest you either try -.0179 or take the given MoI as being about the rod's centre.

 

[p37]

The negative was right. Thank you!